Question

Michael makes 60% of his free-throw attempts. If Michael attempts two free-throws,

what is the probability that he will make exactly one of them? Express your answer as a
common fraction in simplest form.

Answer 1

Answer: 12/25

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Reason:

60% = 60/100 = 0.6 is the probability of making any given free-throw.

1 - 0.6 = 0.4 is the probability of missing any given free-throw.

We have these probabilities

• A = P(making 1st, missing 2nd) = 0.6*0.4 = 0.24
• B = P(missing 1st, making 2nd) = 0.4*0.6 = 0.24

The probability of making exactly one free throw is A+B = 0.24+0.24 = 0.48

Convert this to a fraction:

0.48 = 48/100 = (4*12)/(4*25) = 12/25